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Instructions:
Please read the following instructions carefully before submitting assignment:
It should be clear that your assignment will not get any credit if:

 The assignment is submitted after due date.
 The assignment is submitted by email.
 The submitted assignment does not open or file is corrupt.
 All types of plagiarism are strictly prohibited.
 It is in some format other than .doc (Microsoft Word).

Objectives:
 To learn and understand the basic concepts of I/O bus in CPU.

For any query about the assignment, contact at cs501@vu.edu.pk

Question 1 Marks 10
Problem Statement:
Consider a 16-bit parallel output port attached with the FALCON-A CPU as shown in the figure below:

This port has been mapped onto address DEh of the FALCON-A’s I/O space. To display the data being received from the FALCON-A’s data bus, sixteen (16) LED branches are used. Every LED branch is wired in such a way that when a 1 appears on the particular data bus bit, it turns the LED on; a 0 turns it off.

Question: Which LEDs will be ON when the below given OUT instruction executes on the CPU? Explain in detail.

OUT R2, DEh

Assume that the value of R2 is 1556h.

Question 2 Marks 5

Consider an I/O bus requires 15 nsec for bus requests, 20 nsec for arbitration and the average time to complete an operation is 12 nsec after the access to the bus has been granted, is it possible for such a bus to perform 40 million IOPS?

Note: No assignment will be accepted after due date.

Spring%202015_CS501_3.doc


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Replies to This Discussion

Question # 1 ka koi complete solution de do please....

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Neelam afridi email sharing is not allowed at the site.thanks

Sab log dua me yad rkhna is ku adit kr k de do.......

Attachments:

In second question you have put 10^9.... It should be 10^-9... :)

Q2 ka answer 21.3 hy to ya perform krny k liya possible ho ga ya ni??? 

nahi hoga..

 

which one is the correct splution of question No2

Solution No#1

For the 40million IOPS ,the average time for each IOPS is 1/(40*15)=37 nsec

The sum of the three times is: 15+20+12=42nsec

This mean that bus perform a maximum of: 1/(42*10)=23.9

Thus this is not able to perfom the  40million IOPS .

solution No #2

Answer:

Since for 40 million IOPS, the average time for each IOP is 1 / (40 x 106) =25 nsec. Therefore, the sum of the three times is 15 + 20 + 12 = 47 nsec for a complete I/O operation.

This means that the bus can perform a maximum of 1 / (47 x 109) = 21.3 million IOPS.

 

Solution No. 2

Final Solution:

Q no 1:

Tariq bhai ye last D0 mi 1 kesy aa gyaaaaaaaa????

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